Here is my 'shot' at it
there are
5 x 4 x 3 x 2 x 1 = 120 possible numbers,
so each didgit must appear in each position 120/5 = 24 times
So each position of the 5 digit numbers must sum up to
24(1) + 24(3) + 24(4) + 24(5) + 24(9) = 24(1+3+4+5+9) = 528
P5 P4 P3 P2 P1
528 528 528 528 528
adding these together (P1 would be 8 carry 52 to P2 etc)
I come up with
5,866,608 (I hope it is corrrect !)